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Edexcel International A Level Maths · M2 · Mechanics 2

Projectiles

28 free, high-quality past-paper-style questions on Projectiles for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 260 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-M2-1-B-01 M2 · A2Easy6 marks

A marble rolls off the edge of a horizontal table top with speed 2.5 m s−1. The table top is 0.98 m above a horizontal floor. The marble is modelled as a particle moving freely under gravity from the moment it leaves the table. Take g = 9.8 m s−2.

2.5 m s⁻¹0.98 m
Figure: the marble leaves the edge of the table horizontally; not to scale.
(a)

Find the time taken for the marble to reach the floor.

(2)
(b)

Find the horizontal distance from the edge of the table to the point where the marble hits the floor.

(1)
(c)

Find the speed of the marble as it hits the floor.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
vertically, initial vertical velocity 0: 0.98 = ½ × 9.8 × t2M1
t = √0.2 = 0.447 s (accept 0.45 s)A1
Part (b)
2.5 × 0.447 = 1.12 m (accept 1.1 m; ft their t)B1
Part (c)
vertical component: v = 0 + 9.8 × 0.447 = 4.38 m s−1 (or v2 = 2 × 9.8 × 0.98)M1
speed = √(2.52 + 4.382)M1
= 5.05 m s−1 (accept 5.0)A1

Common error: Giving the marble an initial vertical velocity of 2.5 m s−1. It leaves the table horizontally, so its initial vertical velocity is zero.

2IALM-M2-1-B-02 M2 · A2Easy6 marks

A pebble is flicked from a point O on horizontal ground with speed 12 m s−1 at an angle α above the horizontal, where tan α = 4/3. The pebble is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

12 m s⁻¹αO
Figure: the pebble projected from O at angle α to the horizontal, where tan α = 4/3; not to scale.
(a)

Find the time taken for the pebble to reach its greatest height.

(2)
(b)

Find the greatest height of the pebble above the ground.

(2)
(c)

Find the distance from O to the point where the pebble first hits the ground.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
vertical component 12 × 0.8 = 9.6 m s−1; at the top 0 = 9.6 − 9.8tM1
t = 0.980 s (accept 0.98 s)A1
Part (b)
0 = 9.62 − 2 × 9.8 × hM1
h = 4.70 m (accept 4.7 m)A1
Part (c)
time of flight = 2 × 0.980 = 1.96 s and horizontal component 12 × 0.6 = 7.2 m s−1; distance = 7.2 × 1.96M1
= 14.1 mA1

Exam tip: With tan α = 4/3, draw a 3–4–5 triangle: sin α = 0.8 and cos α = 0.6 exactly, so no angle needs to be found.

3IALM-M2-1-B-03 M2 · A2Easy7 marks

A beanbag is thrown from a point on horizontal ground with speed 14 m s−1 at an angle of 60° above the horizontal. The beanbag is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

14 m s⁻¹60°
Figure: the beanbag thrown at 60° above the horizontal; not to scale.
(a)

State one modelling assumption, other than treating the beanbag as a particle, that has been made.

(1)
(b)

Show that, 2.0 s after it is thrown, the vertical component of the velocity of the beanbag is 7.48 m s−1 downwards, to 3 significant figures, and verify that it is still in the air.

(2)
(c)

Find the speed of the beanbag 2.0 s after it is thrown.

(2)
(d)

Find the direction of motion of the beanbag at this time.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
any one: air resistance is negligible / the only force acting is the weight / g is constantB1
Part (b)
14 sin 60° − 9.8 × 2.0 = 12.12 − 19.6 = −7.48, so 7.48 m s−1 downwardsB1
time of flight 2 × 14 sin 60° ÷ 9.8 = 2.47 s > 2.0 s, so still in the airB1
Part (c)
horizontal component 14 cos 60° = 7 m s−1; speed = √(72 + 7.482)M1
= 10.2 m s−1A1
Part (d)
tan β = 7.48 ÷ 7M1
β = 46.9° below the horizontalA1

Exam tip: A direction of motion needs “above” or “below” the horizontal. The sign of the vertical component tells you which.

4IALM-M2-1-B-04 M2 · A2Easy7 marks

A particle P is projected from a fixed point O with velocity (6i + 8j) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. P moves freely under gravity. Take g = 9.8 m s−2.

(a)

Find the speed of projection and the angle of projection.

(3)
(b)

Find the velocity of P 1.5 s after projection, giving your answer in the form ai + bj.

(2)
(c)

Find the position vector of P relative to O at this time.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
speed = √(62 + 82) = 10 m s−1B1
tan θ = 8 ÷ 6M1
θ = 53.1° above the horizontalA1
Part (b)
v = (6i + 8j) − 9.8 × 1.5jM1
= 6i − 6.7j m s−1A1
Part (c)
r = (6i + 8j) × 1.5 − 4.9 × 1.52jM1
= 9i + 0.975j mA1

Exam tip: In vector form only the j-component is affected by gravity: v = u − gtj and r = ut − ½gt2j.

5IALM-M2-1-FB-01 M2 · A2Easy7 marks

Parcels leave the end E of a horizontal conveyor belt with the speed of the belt and fall into an open bin on a horizontal floor. The diagram shows the dimensions: E is 1.6 m above the floor, and the open top of the bin is 0.375 m above the floor and stretches from 0.9 m to 1.5 m horizontally from E. Each parcel is modelled as a particle moving freely under gravity after it leaves E, and the sides of the bin have negligible thickness. A parcel lands in the bin if, when it has fallen to the level of the top of the bin, its horizontal distance from E is between 0.9 m and 1.5 m. Take g = 9.8 m s−2.

conveyor beltEbin1.6 m0.375 m0.9 m1.5 m
(a)

Show that a parcel takes 0.5 s to fall from E to the level of the top of the bin.

(2)
(b)

Find the range of belt speeds for which the parcels land in the bin.

(3)
(c)

The belt is set to 2.2 m s−1. Find the speed of a parcel as it reaches the level of the top of the bin.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
vertical drop 1.6 − 0.375 = 1.225 m, with zero initial vertical velocity: 1.225 = ½ × 9.8 × t2M1
t2 = 0.25, so t = 0.5 s (given answer)A1
Part (b)
horizontal distance = 0.5u, where u m s−1 is the belt speedM1
0.9 ≤ 0.5u ≤ 1.5M1
1.8 ≤ u ≤ 3.0A1
Part (c)
vertical component 9.8 × 0.5 = 4.9 m s−1; speed = √(2.22 + 4.92)M1
= 5.37 m s−1A1

Common error: Using the full 1.6 m as the drop. The parcel only has to fall to the level of the top of the bin, which is 1.225 m below E.

6IALM-M2-1-FB-02 M2 · A2Easy7 marks

A ball is struck from a point O on horizontal ground. It moves freely under gravity and lands on the ground 3 s later. The graph shows how v m s−1, the vertical component of the velocity of the ball (upwards positive), varies with the time t seconds after the ball is struck. The horizontal component of the velocity is constant at 8 m s−1.

−14.7−9.8−4.904.99.814.70.511.522.530t / sv / m s⁻¹
(a)

Use the graph to show that the acceleration of the ball is 9.8 m s−2 vertically downwards.

(1)
(b)

Use the graph to find the greatest height of the ball above the ground.

(3)
(c)

Find the speed of projection and the angle of projection.

(2)
(d)

Find the distance from O of the point where the ball lands.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
gradient = (−14.7 − 14.7) ÷ 3 = −9.8, so the acceleration is 9.8 m s−2 downwardsB1
Part (b)
greatest height when v = 0, at t = 1.5B1
height = area under the graph from t = 0 to 1.5 = ½ × 1.5 × 14.7M1
= 11.0 m (11.025; accept 11 m)A1
Part (c)
speed = √(82 + 14.72) = 16.7 m s−1B1
angle = tan−1(14.7 ÷ 8) = 61.4° above the horizontalB1
Part (d)
8 × 3 = 24 mB1

Exam tip: A velocity–time graph of the vertical motion works exactly as in straight-line kinematics: the gradient is the acceleration and the area is the vertical displacement.

7IALM-M2-1-FB-03 M2 · A2Easy8 marks

A particle P is projected from a point O on horizontal ground with velocity (ai + bj) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. P moves freely under gravity and lands at the point L on the ground. The diagram shows the path of P on a grid, with distances in metres. A and B are the positions of P 1 s and 2 s after projection, and their coordinates relative to O are shown. Take g = 9.8 m s−2.

24681012141612345678Ox / my / mA (6, 7.1)B (12, 4.4)L
(a)

Use the coordinates of A to find the values of a and b.

(3)
(b)

Find the velocity of P at B, giving your answer in the form pi + qj, and find the speed of P at B.

(3)
(c)

Find the distance OL.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontally 6 m in 1 s, so a = 6B1
vertically: b × 1 − 4.9 × 12 = 7.1M1
b = 12A1
Part (b)
v = (6i + 12j) − 9.8 × 2jM1
= 6i − 7.6j m s−1A1
speed = √(62 + 7.62) = 9.68 m s−1A1
Part (c)
lands when 12t − 4.9t2 = 0, so t = 12 ÷ 4.9 = 2.449 sM1
OL = 6 × 2.449 = 14.7 mA1

Exam tip: Check your values against the second labelled point: with a = 6 and b = 12, the position at t = 2 is (12, 24 − 19.6) = (12, 4.4), which matches B.

8IALM-M2-1-FB-04 M2 · A2Easy7 marks

Two jets of water, A and B, leave a nozzle N at ground level in a garden. The diagram shows their paths, drawn to scale on a grid. Both jets reach the same greatest height, 3.6 m, and they land on the horizontal ground 6 m and 12 m from N respectively, at the grid points shown. Each drop of water is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

1234567891011121234Nhorizontal distance / mheight / m3.6 mjet Ajet B
(a)

Show that, for both jets, the vertical component of the initial velocity is 8.4 m s−1.

(2)
(b)

Find the time of flight of each jet.

(1)
(c)

Find the speed of projection and the angle of projection of each jet.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
at the greatest height the vertical component is zero: 0 = uy2 − 2 × 9.8 × 3.6M1
uy2 = 70.56, so uy = 8.4 (given answer)A1
Part (b)
2 × 8.4 ÷ 9.8 = 1.71 s for both jets (12/7 s)B1
Part (c)
horizontal components: distance ÷ time of flightM1
A: 6 ÷ 1.714 = 3.5 m s−1; B: 12 ÷ 1.714 = 7 m s−1A1
speeds √(3.52 + 8.42) = 9.1 m s−1 and √(72 + 8.42) = 10.9 m s−1A1
angles tan−1(8.4 ÷ 3.5) = 67.4° and tan−1(8.4 ÷ 7) = 50.2°A1

Exam tip: Equal greatest heights mean equal vertical components, and so equal times of flight. Only the horizontal components differ, in the ratio of the ranges.

9IALM-M2-1-B-05 M2 · A2Medium9 marks

A ball is thrown horizontally from an open window. The point of release is 12.5 m vertically above horizontal ground. The ball hits the ground at a point which is 18 m horizontally from the point of release. The ball is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

12.5 m18 m
Figure: the ball is thrown horizontally from the window and lands 18 m from the point of release, measured horizontally; not to scale.
(a)

Find the time taken for the ball to reach the ground.

(2)
(b)

Find the speed with which the ball is thrown.

(2)
(c)

Find the speed of the ball as it hits the ground.

(3)
(d)

Find the angle between the direction of motion of the ball and the ground at this instant.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
12.5 = ½ × 9.8 × t2M1
t = 1.60 s (1.597)A1
Part (b)
u = 18 ÷ 1.597M1
= 11.3 m s−1A1
Part (c)
vertical component 9.8 × 1.597 = 15.65 m s−1M1
speed = √(11.272 + 15.652)M1
= 19.3 m s−1A1
Part (d)
tan β = 15.65 ÷ 11.27M1
β = 54.2° (accept 54°)A1

Exam tip: Keep the unrounded time (1.597… s) in your calculator. Using 1.6 s in later parts can change the third significant figure.

10IALM-M2-1-B-06 M2 · A2Medium7 marks

A stone is thrown from a point A on a bridge. A is 22 m vertically above the surface of a river. The stone is thrown with speed 8 m s−1 at an angle of 20° below the horizontal. The stone moves freely under gravity until it hits the water. Take g = 9.8 m s−2.

river8 m s⁻¹20°A22 m
Figure: the stone thrown from A at 20° below the horizontal; not to scale.
(a)

Find the time taken for the stone to reach the water.

(4)
(b)

Find the horizontal distance travelled by the stone in this time.

(2)
(c)

Explain why only one root of your quadratic equation in part (a) is used.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
downwards positive: initial vertical component 8 sin 20° = 2.736 m s−1B1
22 = 2.736t + 4.9t2M1
solving the quadratic: t = (−2.736 + √(2.7362 + 4 × 4.9 × 22)) ÷ 9.8M1
t = 1.86 sA1
Part (b)
8 cos 20° × 1.858M1
= 14.0 m (accept 14 m)A1
Part (c)
the other root is negative (−2.42), which would be a time before the stone was thrownB1

Common error: Using +8 sin 20° as an upward component when the stone is thrown below the horizontal. Draw the velocity first and decide which way the vertical component points.

11IALM-M2-1-B-07 M2 · A2Medium8 marks

In a game, a dart is thrown horizontally towards a board. The dart is released at a horizontal distance of 2.4 m from the board, level with the centre of the board, and it hits the board at a point 0.12 m vertically below the centre. The dart is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

centrerelease2.4 m0.12 m
Figure: the dart's path from the point of release to the board (drop exaggerated); not to scale.
(a)

Find the time taken for the dart to reach the board.

(2)
(b)

Find the speed with which the dart is thrown.

(2)
(c)

The player throws a second dart horizontally from the same point. It hits the board 0.05 m below the centre. Find the speed of this throw.

(3)
(d)

In practice, air resistance acts on the dart. State, with a reason, how this would affect your answer to part (b).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
0.12 = ½ × 9.8 × t2M1
t = 0.156 sA1
Part (b)
u = 2.4 ÷ 0.1565M1
= 15.3 m s−1A1
Part (c)
0.05 = 4.9t2M1
t = 0.101 sA1
u = 2.4 ÷ 0.101 = 23.8 m s−1A1
Part (d)
the true speed of throw would need to be greater (the answer is an underestimate), because air resistance slows the dart horizontally so it takes longer to reach the board and drops furtherB1

Exam tip: For a horizontal throw, the drop fixes the time (vertical motion only), and the time then fixes the speed (horizontal motion only).

12IALM-M2-1-B-08 M2 · A2Medium9 marks

A machine on horizontal ground fires tennis balls with speed 16 m s−1 at an angle θ above the horizontal. The balls are modelled as particles moving freely under gravity, and the balls leave the machine at ground level. Take g = 9.8 m s−2.

(a)

Show that the horizontal distance R m travelled by a ball before it lands is given by R = 256/9.8 sin 2θ.

(4)
(b)

Find the two values of θ for which a ball lands 20 m from the machine.

(3)
(c)

Find the time of flight for the larger value of θ.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
vertically: 0 = 16 sin θ t − 4.9t2M1
t = 32 sin θ ÷ 9.8 (non-zero root)A1
R = 16 cos θ × 32 sin θ ÷ 9.8M1
= 256 × 2 sin θ cos θ ÷ 9.8 = 256/9.8 sin 2θ (given answer, using sin 2θ = 2 sin θ cos θ)A1
Part (b)
sin 2θ = 20 × 9.8 ÷ 256 = 0.7656M1
2θ = 49.96° or 130.04°M1
θ = 25.0° or 65.0°A1
Part (c)
t = 32 sin 65.0° ÷ 9.8M1
= 2.96 sA1

Exam tip: sin 2θ = k has two solutions with 0° < 2θ < 180°. Halving them gives two angles of projection that add to 90°.

13IALM-M2-1-B-09 M2 · A2Medium9 marks

A ball is thrown from a point O with speed 13 m s−1 at an angle θ above the horizontal, where tan θ = 12/5. The ball moves freely under gravity in a vertical plane. Take O as the origin, with x m and y m the horizontal and vertically upward displacements of the ball from O. Take g = 9.8 m s−2.

5 m above O13 m s⁻¹θO
Figure: the ball thrown from O, with the balcony level 5 m above O; tan θ = 12/5; not to scale.
(a)

Show that the equation of the path of the ball is y = 2.4x − 0.196x2.

(4)
(b)

A friend on a balcony catches the ball at a point 5 m above the level of O while the ball is falling. Find the horizontal distance from O to this point.

(3)
(c)

Find the time between the ball first rising through the level of the balcony and being caught.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
sin θ = 12/13, cos θ = 5/13, so components 5 m s−1 and 12 m s−1B1
x = 5t, so t = x/5M1
y = 12(x/5) − 4.9(x/5)2M1
y = 2.4x − 0.196x2 (given answer)A1
Part (b)
0.196x2 − 2.4x + 5 = 0M1
x = (2.4 ± √(2.42 − 4 × 0.196 × 5)) ÷ 0.392 = 2.66 or 9.58M1
falling, so the larger root: 9.58 mA1
Part (c)
time = (9.583 − 2.662) ÷ 5M1
= 1.38 sA1

Exam tip: The trajectory equation gives positions, not times. Divide a horizontal distance by the constant horizontal velocity to turn it into a time.

14IALM-M2-1-B-10 M2 · A2Medium9 marks

A fountain nozzle at ground level projects drops of water with speed 11 m s−1 at an angle of 70° above the horizontal. Each drop is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

(a)

Find the horizontal and vertical components of the initial velocity of a drop.

(2)
(b)

Find the two times after projection at which a drop is moving at 45° to the horizontal.

(4)
(c)

Find the height of the drop above the ground at the first of these times.

(2)
(d)

Without further calculation, state the height of the drop at the second time, giving a reason.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
11 cos 70° = 3.76 m s−1B1
11 sin 70° = 10.3 m s−1 (10.34)B1
Part (b)
at 45°, the vertical component has the same size as the horizontal component, 3.762 m s−1M1
rising: 10.34 − 9.8t = 3.762M1
t = 0.671 sA1
falling: 10.34 − 9.8t = −3.762, t = 1.44 sA1
Part (c)
y = 10.34 × 0.6709 − 4.9 × 0.67092M1
= 4.73 mA1
Part (d)
4.73 m, because the two times are equally spaced either side of the time of the top and the path is symmetricalB1

Exam tip: Moving at 45° means |vvertical| = vhorizontal. There are two such moments: one on the way up and one on the way down.

15IALM-M2-1-FB-05 M2 · A2Medium9 marks

A ball is kicked from a point O on horizontal ground with speed u m s−1 at an angle θ above the horizontal, and moves freely under gravity. The graph shows the path of the ball, where x m and y m are its horizontal and vertical distances from O. The highest point of the path is at a grid point, and the ball lands at x = 20. A vertical wall of height 3 m stands at x = 16. Take g = 9.8 m s−2.

2468101214161820123456Ox / my / mwallhighest point
(a)

Write down the greatest height of the ball and its horizontal distance from O at that moment. Hence show that tan θ = 1.

(3)
(b)

Find the value of u.

(2)
(c)

Show that the equation of the path is y = x − 0.05x2.

(2)
(d)

Determine whether the ball passes over the wall and, if so, by how much it clears it.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
greatest height 5 m, at x = 10B1
if t1 is the time to the top: 5 = ½ u sin θ t1 and 10 = u cos θ t1M1
dividing: ½ tan θ = ½, so tan θ = 1 (given answer)A1
Part (b)
(u sin 45°)2 = 2 × 9.8 × 5 = 98M1
u2 = 196, u = 14A1
Part (c)
y = x tan 45° − 9.8x2 ÷ (2 × 142 × cos2 45°)M1
= x − 9.8x2 ÷ 196 = x − 0.05x2 (given answer)A1
Part (d)
at x = 16: y = 16 − 0.05 × 256 = 3.2M1
3.2 > 3, so it clears the wall by 0.2 mA1

Exam tip: Use the graph for values at grid points only. A clearance of 0.2 m is too small to read from a diagram, so calculate it from the equation of the path.

16IALM-M2-1-FB-06 M2 · A2Medium9 marks

A ball is kicked from a point O on horizontal ground with speed u m s−1 at 30° above the horizontal. It moves freely under gravity and lands at the point T on the flat top of a platform. The diagram shows the dimensions: the platform is 1.5 m high, its front edge is 15 m horizontally from O, and T is 18 m horizontally from O. Take g = 9.8 m s−2.

platformOT30°15 m18 m1.5 m
(a)

Find the value of u.

(4)
(b)

Find the time taken for the ball to reach T.

(2)
(c)

Show that the ball is moving downwards when it reaches T.

(2)
(d)

Verify that the ball passes above the front edge of the platform.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
using y = x tan θ − gx2 ÷ (2u2 cos2 θ) or eliminating tM1
1.5 = 18 tan 30° − 9.8 × 182 ÷ (2u2 × 0.75)A1
u2 = 3175.2 ÷ (1.5 × (10.392 − 1.5)) = 238.0M1
u = 15.4A1
Part (b)
t = 18 ÷ (15.43 cos 30°)M1
= 1.35 sA1
Part (c)
vertical component = 15.43 sin 30° − 9.8 × 1.347M1
= −5.49 < 0, so the ball is moving downwardsA1
Part (d)
at x = 15: y = 15 tan 30° − 9.8 × 225 ÷ (2 × 238.0 × 0.75) = 2.49 > 1.5, so it passes above the edgeB1

Exam tip: When the speed is the unknown, the equation of the path, y = x tan θ − gx2/(2u2 cos2 θ), gives u2 directly without finding the time first.

17IALM-M2-1-FB-07 M2 · A2Medium9 marks

A ball is projected from a point O on horizontal ground and moves freely under gravity. The diagram shows its path. At the point Q, 1.5 s after projection, the ball is moving with speed 10 m s−1 in the direction shown, 30° below the horizontal. Take g = 9.8 m s−2.

O30°Q10 m s⁻¹
(a)

Find the horizontal and vertical components of the initial velocity of the ball.

(3)
(b)

Find the speed and the angle of projection.

(3)
(c)

Find the height of Q above the ground.

(2)
(d)

Find how long after passing through Q the ball lands.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontal component = 10 cos 30° = 8.66 m s−1 (5√3)B1
vertically, upwards positive: −10 sin 30° = uy − 9.8 × 1.5M1
uy = −5 + 14.7 = 9.7 m s−1A1
Part (b)
speed = √(8.662 + 9.72)M1
= 13.0 m s−1A1
angle = tan−1(9.7 ÷ 8.66) = 48.2° above the horizontalA1
Part (c)
9.7 × 1.5 − 4.9 × 1.52M1
= 3.525 m (accept 3.52 or 3.53)A1
Part (d)
time of flight 2 × 9.7 ÷ 9.8 = 1.980 s, so 1.980 − 1.5 = 0.480 sB1

Common error: Taking the vertical component at Q as +5 m s−1. The ball is moving 30° below the horizontal, so with upwards positive the component is −5 m s−1.

18IALM-M2-1-FB-08 M2 · A2Medium9 marks

A ball is thrown from a point A, 2.1 m above horizontal ground, and moves freely under gravity until it hits the ground. The graphs show x m, the horizontal displacement of the ball from A, and y m, the height of the ball above the ground, against the time t seconds after it is thrown. The labelled points give exact values. Take g = 9.8 m s−2.

00.511.522.505101520t / sx / m(2, 15)00.511.522.5012345678t / sy / m(0, 2.1)(1.0, 7.0)
(a)

Use the graphs to find the horizontal and vertical components of the initial velocity of the ball.

(3)
(b)

Find the speed and the angle of projection.

(2)
(c)

Find the time at which the ball hits the ground.

(3)
(d)

Find the horizontal distance travelled by the ball before it hits the ground.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontal component = 15 ÷ 2 = 7.5 m s−1B1
the highest point is at t = 1.0, so 0 = uy − 9.8 × 1.0 (or 7.0 − 2.1 = uy2 ÷ 19.6)M1
uy = 9.8 m s−1A1
Part (b)
speed = √(7.52 + 9.82) = 12.3 m s−1B1
angle = tan−1(9.8 ÷ 7.5) = 52.6° above the horizontalB1
Part (c)
displacement −2.1 m: −2.1 = 9.8t − 4.9t2M1
4.9t2 − 9.8t − 2.1 = 0, t = (9.8 + √(9.82 + 4 × 4.9 × 2.1)) ÷ 9.8M1
t = 2.20 s (2.195)A1
Part (d)
7.5 × 2.195 = 16.5 mB1

Exam tip: Read values only at labelled points or grid crossings. The landing time is not on a gridline here, so it has to be calculated from the vertical motion.

19IALM-M2-1-FB-09 M2 · A2Medium10 marks

A ball is thrown from the edge E of the flat roof of building A with speed 12 m s−1 at 25° above the horizontal. The diagram shows the dimensions: E is 20 m above horizontal ground, and building B has a flat roof 8 m above the ground that begins 10 m horizontally from E and is 15 m long. The ball moves freely under gravity in the vertical plane shown. Take g = 9.8 m s−2.

ABE25°12 m s⁻¹20 m8 m10 m15 m
(a)

Find the time taken for the ball to fall to the level of the roof of B.

(4)
(b)

Show that the ball lands on the roof of B, and find its distance from the near edge of that roof.

(3)
(c)

Find the speed of the ball as it lands.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
vertical component of initial velocity 12 sin 25° = 5.071 m s−1B1
displacement 8 − 20 = −12 m: −12 = 5.071t − 4.9t2M1
t = (5.071 + √(5.0712 + 4 × 4.9 × 12)) ÷ 9.8M1
t = 2.17 s (2.166)A1
Part (b)
horizontal distance = 12 cos 25° × 2.166M1
= 23.6 mA1
10 < 23.6 < 25, so it lands on the roof of B, 13.6 m from the near edgeA1
Part (c)
vertical component 5.071 − 9.8 × 2.166 = −16.15 m s−1M1
speed = √(10.882 + 16.152) (or √(122 + 2 × 9.8 × 12))M1
= 19.5 m s−1A1

Exam tip: Measure every vertical displacement from the point of projection. Landing on a roof 8 m high, starting from 20 m, is a displacement of −12 m.

20IALM-M2-1-FB-10 M2 · A2Medium9 marks

A stunt rider leaves the lip L of a ramp with speed u m s−1 at 20° above the horizontal. The diagram shows the landing deck, whose horizontal top is level with L. The near end of the deck is 9 m horizontally from L and the deck is 6 m long. The rider and bike are modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

landing deckLu m s⁻¹20°9 m6 mgap
(a)

Find the range of values of u for which the rider lands on the deck.

(5)
(b)

Given that u = 13, find the greatest height of the rider above the level of L.

(2)
(c)

Given that u = 13, find the distance from the near end of the deck to the point where the rider lands.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
time to return to the level of L: t = 2u sin 20° ÷ 9.8M1
horizontal distance = u cos 20° × 2u sin 20° ÷ 9.8 = u2 sin 40° ÷ 9.8A1
9 ≤ u2 sin 40° ÷ 9.8 ≤ 15M1
lower limit u ≥ 11.7A1
upper limit u ≤ 15.1, so 11.7 ≤ u ≤ 15.1A1
Part (b)
(13 sin 20°)2 ÷ (2 × 9.8)M1
= 1.01 mA1
Part (c)
132 sin 40° ÷ 9.8 = 11.08 m from LM1
11.08 − 9 = 2.08 m (accept 2.1 m)A1

Common error: Using only the 9 m gap. “Lands on the deck” has a far end too, so there is an upper limit on the speed as well as a lower one.

21IALM-M2-1-B-11 M2 · A2Hard12 marks

A water balloon is launched from a point O with speed 24 m s−1 at an angle θ above the horizontal. It passes through an open window W. The point W is 36 m horizontally from O and 8.5 m vertically above the level of O. The balloon is modelled as a particle moving freely under gravity. Take g = 9.8 m s−2.

W24 m s⁻¹θO8.5 m36 m
Figure: the launch point O and the window W; the path of the balloon is not shown; not to scale.
(a)

Show that 441 tan2 θ − 1440 tan θ + 781 = 0.

(5)
(b)

Hence find the two possible values of θ.

(3)
(c)

For the smaller value of θ, find the time taken for the balloon to reach W.

(2)
(d)

For the smaller value of θ, determine whether the balloon is moving upwards or downwards as it passes through W.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontally: 36 = 24 cos θ × t, so t = 1.5/cos θM1
vertically: 8.5 = 24 sin θ × t − 4.9t2M1
8.5 = 36 tan θ − 11.025 sec2 θA1
use of sec2 θ = 1 + tan2 θ: 8.5 = 36 tan θ − 11.025 − 11.025 tan2 θM1
× 40: 441 tan2 θ − 1440 tan θ + 781 = 0 (given answer)A1
Part (b)
tan θ = (1440 ± √(14402 − 4 × 441 × 781)) ÷ 882M1
tan θ = 0.6868 or 2.578A1
θ = 34.5° or 68.8°A1
Part (c)
t = 36 ÷ (24 cos 34.48°)M1
= 1.82 sA1
Part (d)
vertical velocity = 24 sin 34.48° − 9.8 × 1.820M1
= −4.25 m s−1, negative, so it is moving downwardsA1

Exam tip: Replace t using the horizontal equation, write 1/cos2 θ as sec2 θ, then use sec2 θ = 1 + tan2 θ. Show each of these lines in a “show that”.

22IALM-M2-1-B-12 M2 · A2Hard13 marks

A particle P is projected from a point A with velocity (8i + 10j) m s−1, where i and j are unit vectors horizontally and vertically upwards respectively. The point O is on horizontal ground and the position vector of A relative to O is (2i + 3j) m. P moves freely under gravity until it hits the ground. Take g = 9.8 m s−2.

OA3 m2 m(8i + 10j) m s⁻¹
Figure: P is projected from A, whose position relative to O is (2i + 3j) m; not to scale.
(a)

Find the position vector of P relative to O at time t seconds after projection.

(2)
(b)

Find the time taken for P to reach the ground.

(3)
(c)

Find the distance from O of the point where P hits the ground.

(2)
(d)

Find the greatest height of P above the ground, and its speed at that instant.

(3)
(e)

Find the time at which P is moving in a direction perpendicular to its initial direction of motion.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
r = (2i + 3j) + (8i + 10j)t − 4.9t2jM1
r = (2 + 8t)i + (3 + 10t − 4.9t2)jA1
Part (b)
3 + 10t − 4.9t2 = 0M1
t = (10 + √(100 + 58.8)) ÷ 9.8M1
= 2.31 s (2.306)A1
Part (c)
2 + 8 × 2.306M1
= 20.5 mA1
Part (d)
at the top the j-component of velocity is zero: 0 = 102 − 2 × 9.8 × s, s = 5.10M1
height = 3 + 5.10 = 8.10 mA1
speed = 8 m s−1 (the i-component only)B1
Part (e)
v = 8i + (10 − 9.8t)j; perpendicular to 8i + 10j when the scalar product is zero, or gradient −0.8M1
64 + 10(10 − 9.8t) = 0M1
t = 164 ÷ 98 = 1.67 s, which is before P landsA1

Exam tip: If the particle does not start at O, add the starting position vector: r = r0 + ut − ½gt2j.

23IALM-M2-1-B-13 M2 · A2Hard13 marks

A stone is catapulted from a point C on the top of a vertical castle wall. C is 30 m above horizontal ground. The stone is projected with speed 24 m s−1 at an angle of 40° above the horizontal and moves freely under gravity until it hits the ground. Take g = 9.8 m s−2.

24 m s⁻¹40°C30 m
Figure: the stone catapulted from C on top of the wall; not to scale.
(a)

Find the greatest height of the stone above the ground.

(3)
(b)

Find the time taken for the stone to reach the ground.

(4)
(c)

Find the horizontal distance from the foot of the wall to the point where the stone lands.

(2)
(d)

Find the length of time for which the stone is more than 35 m above the ground.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
vertical component 24 sin 40° = 15.43 m s−1; 0 = 15.432 − 2 × 9.8 × sM1
s = 12.14 mA1
height above ground = 30 + 12.14 = 42.1 mA1
Part (b)
upwards positive, displacement −30B1
−30 = 15.43t − 4.9t2M1
t = (15.43 + √(15.432 + 4 × 4.9 × 30)) ÷ 9.8M1
= 4.51 s (4.507)A1
Part (c)
24 cos 40° × 4.507 = 18.39 × 4.507M1
= 82.9 mA1
Part (d)
35 m above ground is 5 m above C: 5 = 15.43t − 4.9t2M1
solving 4.9t2 − 15.43t + 5 = 0: t = (15.43 ± √(15.432 − 4 × 4.9 × 5)) ÷ 9.8M1
t = 0.367 or 2.78A1
time = 2.781 − 0.367 = 2.41 sA1

Common error: Using the level-ground formula for the range when the stone lands below its launch point. Find the time from the vertical quadratic first, then use the horizontal motion.

24IALM-M2-1-B-14 M2 · A2Hard12 marks

A particle is projected from a point O on horizontal ground with speed u m s−1 at an angle θ above the horizontal. It moves freely under gravity, where the acceleration due to gravity is g m s−2.

(a)

Show that the time taken for the particle to return to the ground is 2u sin θ/g.

(3)
(b)

Hence show that the horizontal distance travelled in this time is u2 sin 2θ/g.

(2)
(c)

A long-jumper takes off with speed 8.5 m s−1 and is modelled as a particle moving freely under gravity, taking off and landing on the same horizontal level. Using g = 9.8, find the two possible angles of take-off that give a jump of 6.4 m.

(3)
(d)

Find the greatest height above the take-off level reached by the long-jumper for the smaller of these angles.

(3)
(e)

State one way in which the particle model is unrealistic for a long-jumper.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
vertically: 0 = (u sin θ)t − ½gt2M1
t(u sin θ − ½gt) = 0M1
t ≠ 0, so t = 2u sin θ/g (given answer)A1
Part (b)
x = u cos θ × 2u sin θ/gM1
= u2(2 sin θ cos θ)/g = u2 sin 2θ/g (given answer)A1
Part (c)
6.4 = 8.52 sin 2θ ÷ 9.8M1
sin 2θ = 0.8681, 2θ = 60.24° or 119.76°M1
θ = 30.1° or 59.9°A1
Part (d)
vertical component 8.5 sin 30.12° = 4.265 m s−1M1
0 = 4.2652 − 2 × 9.8 × hM1
h = 0.928 m (accept 0.93 m)A1
Part (e)
any one: the jumper has size and changes body shape, so cannot be a single point / the centre of mass does not start and end at ground level / air resistance actsB1

Exam tip: In a “show that” with a given answer, every step must be written down, including the identity sin 2θ = 2 sin θ cos θ.

25IALM-M2-1-FB-11 M2 · A2Hard12 marks

A ball is projected from a point O on horizontal ground with speed 15 m s−1 at an angle θ above the horizontal, and moves freely under gravity. The diagram shows a vertical post standing on the ground; its top P is 12 m horizontally from O and 4 m above the ground. The ball must pass above P. Take g = 9.8 m s−2.

PO15 m s⁻¹θ12 m4 m
(a)

Show that the ball passes above P when 98 tan2 θ − 375 tan θ + 223 < 0.

(5)
(b)

Hence find the range of values of θ for which the ball passes above P.

(4)
(c)

For θ = 45°, find the height by which the ball clears P and the distance beyond the post at which the ball lands.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontally 12 = 15 cos θ t, so t = 12 ÷ (15 cos θ)M1
height at the post: y = 12 tan θ − 4.9 × 144 ÷ (225 cos2 θ) = 12 tan θ − 3.136 sec2 θA1
sec2 θ = 1 + tan2 θM1
12 tan θ − 3.136(1 + tan2 θ) > 4M1
3.136 tan2 θ − 12 tan θ + 7.136 < 0; × 125/4: 98 tan2 θ − 375 tan θ + 223 < 0 (given answer)A1
Part (b)
critical values tan θ = (375 ± √(3752 − 4 × 98 × 223)) ÷ 196M1
tan θ = 0.736 or 3.09A1
quadratic negative between the roots: 0.736 < tan θ < 3.09M1
36.4° < θ < 72.1°A1
Part (c)
y = 12 − 3.136 × 2 = 5.728M1
clears P by 1.73 mA1
range 152 ÷ 9.8 = 22.96 m, so it lands 11.0 m beyond the postB1

Exam tip: Write the condition as a quadratic inequality in tan θ. With a positive coefficient of tan2 θ, “&lt; 0” means between the two roots.

26IALM-M2-1-FB-12 M2 · A2Hard11 marks

The diagram shows a point O on horizontal ground and a vertical tower whose foot F is 20 m from O; the top T of the tower is 15 m above F. At time t = 0, a particle A is projected from O with velocity (10i + 15j) m s−1, and at the same instant a particle B is projected horizontally from T towards O with speed V m s−1. Here i and j are unit vectors horizontally (in the direction from O to F) and vertically upwards. Both particles move freely under gravity in the vertical plane through O, F and T. Take g = 9.8 m s−2.

TFOA: (10i + 15j) m s⁻¹B: V m s⁻¹20 m15 mij
(a)

Write down the position vectors of A and of B, relative to O, at time t seconds.

(3)
(b)

Given that A and B collide, show that they collide 1 s after projection, and find V.

(3)
(c)

Find the position vector of the point where they collide.

(1)
(d)

Find the velocities of A and B just before they collide, and the angle between their directions of motion at that instant.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
rA = 10ti + (15t − 4.9t2)jB1
rB = (20 − Vt)i …B1
… + (15 − 4.9t2)jB1
Part (b)
equating j-components: 15t − 4.9t2 = 15 − 4.9t2M1
15t = 15, so t = 1 (given answer)A1
equating i-components at t = 1: 10 = 20 − V, V = 10A1
Part (c)
10i + 10.1j mB1
Part (d)
vA = 10i + (15 − 9.8)j = 10i + 5.2j m s−1B1
vB = −10i − 9.8j m s−1B1
cos φ = (−100 − 50.96) ÷ (11.27 × 14.00)M1
φ = 163°A1

Exam tip: Two particles collide when both components of their position vectors are equal at the same time. Here the gravity terms cancel in the j-components, so t comes out without knowing V.

27IALM-M2-1-FB-13 M2 · A2Hard13 marks

A ball is projected from a point A at the top of a vertical cliff with speed u m s−1 at an angle α above the horizontal, where tan α = 3/4. The ball moves freely under gravity and hits the sea at the point B, 5 s after projection. The diagram shows that A is h m above the sea and that B is 60 m horizontally from A. Take g = 9.8 m s−2.

seaABu m s⁻¹αh m60 m
(a)

Find the value of u.

(2)
(b)

Find the value of h.

(3)
(c)

Find the speed of the ball as it hits the sea, and its direction of motion at that instant.

(4)
(d)

Find the length of time for which the speed of the ball is less than 13 m s−1.

(4)
Show mark scheme
AnswerMarkNotes
Part (a)
cos α = 0.8: u × 0.8 × 5 = 60M1
u = 15A1
Part (b)
vertical component 15 × 0.6 = 9 m s−1B1
−h = 9 × 5 − 4.9 × 52M1
h = 77.5A1
Part (c)
vertical component 9 − 9.8 × 5 = −40 m s−1M1
speed = √(122 + 402) = 41.8 m s−1A1
tan β = 40 ÷ 12M1
β = 73.3° below the horizontalA1
Part (d)
speed2 = 122 + vy2 < 132M1
|vy| < 5A1
−5 < 9 − 9.8t < 5, so 0.408 < t < 1.43M1
length of time = 10 ÷ 9.8 = 1.02 sA1

Exam tip: The horizontal component never changes, so the speed is least at the top of the path (12 m s−1 here). A condition on the speed is really a condition on the vertical component.

28IALM-M2-1-FB-14 M2 · A2Hard13 marks

In a sports hall, a ball is launched from a point O on the floor with speed 21 m s−1 at an angle θ above the horizontal. It moves freely under gravity and is aimed at the point P on the edge of a balcony. The diagram shows that P is 30 m horizontally from O and 5 m above the floor, and that the horizontal ceiling is 12 m above the floor. Take g = 9.8 m s−2.

ceilingbalconyPO21 m s⁻¹θ12 m30 m5 m
(a)

Show that, if the ball passes through P, then 2 tan2 θ − 6 tan θ + 3 = 0.

(5)
(b)

Find the exact values of tan θ, and hence the two possible angles of projection.

(3)
(c)

By finding the greatest height of the ball for each angle, show that only one of these angles can be used.

(3)
(d)

For the angle that can be used, find the time taken for the ball to reach P.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
horizontally t = 30 ÷ (21 cos θ), substituted into the vertical equationM1
5 = 30 tan θ − 4.9 × 900 ÷ (441 cos2 θ)A1
4.9 × 900 ÷ 441 = 10 and sec2 θ = 1 + tan2 θM1
5 = 30 tan θ − 10 − 10 tan2 θA1
10 tan2 θ − 30 tan θ + 15 = 0, so 2 tan2 θ − 6 tan θ + 3 = 0 (given answer)A1
Part (b)
tan θ = (6 ± √12) ÷ 4M1
= (3 ± √3)/2A1
θ = 32.4° or 67.1°A1
Part (c)
greatest height = (21 sin θ)2 ÷ (2 × 9.8)M1
for 32.4°: 6.45 m; for 67.1°: 19.1 mA1
19.1 > 12 and that path reaches its top (16.1 m from O) before P, so it would hit the ceiling; only 32.4° can be used (6.45 < 12)A1
Part (d)
t = 30 ÷ (21 cos 32.37°)M1
= 1.69 sA1

Common error: Giving both roots of the tan θ quadratic as answers. Each root is a possible path through P; check each one against the setting, here the ceiling.